3.2304 \(\int (a+b \sqrt [3]{x})^3 x^3 \, dx\)

Optimal. Leaf size=47 \[ \frac{9}{13} a^2 b x^{13/3}+\frac{a^3 x^4}{4}+\frac{9}{14} a b^2 x^{14/3}+\frac{b^3 x^5}{5} \]

[Out]

(a^3*x^4)/4 + (9*a^2*b*x^(13/3))/13 + (9*a*b^2*x^(14/3))/14 + (b^3*x^5)/5

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Rubi [A]  time = 0.0329955, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac{9}{13} a^2 b x^{13/3}+\frac{a^3 x^4}{4}+\frac{9}{14} a b^2 x^{14/3}+\frac{b^3 x^5}{5} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^3*x^3,x]

[Out]

(a^3*x^4)/4 + (9*a^2*b*x^(13/3))/13 + (9*a*b^2*x^(14/3))/14 + (b^3*x^5)/5

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \left (a+b \sqrt [3]{x}\right )^3 x^3 \, dx &=3 \operatorname{Subst}\left (\int x^{11} (a+b x)^3 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (a^3 x^{11}+3 a^2 b x^{12}+3 a b^2 x^{13}+b^3 x^{14}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{a^3 x^4}{4}+\frac{9}{13} a^2 b x^{13/3}+\frac{9}{14} a b^2 x^{14/3}+\frac{b^3 x^5}{5}\\ \end{align*}

Mathematica [A]  time = 0.0233048, size = 47, normalized size = 1. \[ \frac{9}{13} a^2 b x^{13/3}+\frac{a^3 x^4}{4}+\frac{9}{14} a b^2 x^{14/3}+\frac{b^3 x^5}{5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^3*x^3,x]

[Out]

(a^3*x^4)/4 + (9*a^2*b*x^(13/3))/13 + (9*a*b^2*x^(14/3))/14 + (b^3*x^5)/5

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Maple [A]  time = 0.001, size = 36, normalized size = 0.8 \begin{align*}{\frac{{a}^{3}{x}^{4}}{4}}+{\frac{9\,b{a}^{2}}{13}{x}^{{\frac{13}{3}}}}+{\frac{9\,{b}^{2}a}{14}{x}^{{\frac{14}{3}}}}+{\frac{{b}^{3}{x}^{5}}{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^(1/3))^3*x^3,x)

[Out]

1/4*a^3*x^4+9/13*a^2*b*x^(13/3)+9/14*a*b^2*x^(14/3)+1/5*b^3*x^5

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Maxima [B]  time = 0.976755, size = 270, normalized size = 5.74 \begin{align*} \frac{{\left (b x^{\frac{1}{3}} + a\right )}^{15}}{5 \, b^{12}} - \frac{33 \,{\left (b x^{\frac{1}{3}} + a\right )}^{14} a}{14 \, b^{12}} + \frac{165 \,{\left (b x^{\frac{1}{3}} + a\right )}^{13} a^{2}}{13 \, b^{12}} - \frac{165 \,{\left (b x^{\frac{1}{3}} + a\right )}^{12} a^{3}}{4 \, b^{12}} + \frac{90 \,{\left (b x^{\frac{1}{3}} + a\right )}^{11} a^{4}}{b^{12}} - \frac{693 \,{\left (b x^{\frac{1}{3}} + a\right )}^{10} a^{5}}{5 \, b^{12}} + \frac{154 \,{\left (b x^{\frac{1}{3}} + a\right )}^{9} a^{6}}{b^{12}} - \frac{495 \,{\left (b x^{\frac{1}{3}} + a\right )}^{8} a^{7}}{4 \, b^{12}} + \frac{495 \,{\left (b x^{\frac{1}{3}} + a\right )}^{7} a^{8}}{7 \, b^{12}} - \frac{55 \,{\left (b x^{\frac{1}{3}} + a\right )}^{6} a^{9}}{2 \, b^{12}} + \frac{33 \,{\left (b x^{\frac{1}{3}} + a\right )}^{5} a^{10}}{5 \, b^{12}} - \frac{3 \,{\left (b x^{\frac{1}{3}} + a\right )}^{4} a^{11}}{4 \, b^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3*x^3,x, algorithm="maxima")

[Out]

1/5*(b*x^(1/3) + a)^15/b^12 - 33/14*(b*x^(1/3) + a)^14*a/b^12 + 165/13*(b*x^(1/3) + a)^13*a^2/b^12 - 165/4*(b*
x^(1/3) + a)^12*a^3/b^12 + 90*(b*x^(1/3) + a)^11*a^4/b^12 - 693/5*(b*x^(1/3) + a)^10*a^5/b^12 + 154*(b*x^(1/3)
+ a)^9*a^6/b^12 - 495/4*(b*x^(1/3) + a)^8*a^7/b^12 + 495/7*(b*x^(1/3) + a)^7*a^8/b^12 - 55/2*(b*x^(1/3) + a)^
6*a^9/b^12 + 33/5*(b*x^(1/3) + a)^5*a^10/b^12 - 3/4*(b*x^(1/3) + a)^4*a^11/b^12

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Fricas [A]  time = 1.45517, size = 96, normalized size = 2.04 \begin{align*} \frac{1}{5} \, b^{3} x^{5} + \frac{9}{14} \, a b^{2} x^{\frac{14}{3}} + \frac{9}{13} \, a^{2} b x^{\frac{13}{3}} + \frac{1}{4} \, a^{3} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3*x^3,x, algorithm="fricas")

[Out]

1/5*b^3*x^5 + 9/14*a*b^2*x^(14/3) + 9/13*a^2*b*x^(13/3) + 1/4*a^3*x^4

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Sympy [A]  time = 2.18122, size = 42, normalized size = 0.89 \begin{align*} \frac{a^{3} x^{4}}{4} + \frac{9 a^{2} b x^{\frac{13}{3}}}{13} + \frac{9 a b^{2} x^{\frac{14}{3}}}{14} + \frac{b^{3} x^{5}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/3))**3*x**3,x)

[Out]

a**3*x**4/4 + 9*a**2*b*x**(13/3)/13 + 9*a*b**2*x**(14/3)/14 + b**3*x**5/5

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Giac [A]  time = 1.12135, size = 47, normalized size = 1. \begin{align*} \frac{1}{5} \, b^{3} x^{5} + \frac{9}{14} \, a b^{2} x^{\frac{14}{3}} + \frac{9}{13} \, a^{2} b x^{\frac{13}{3}} + \frac{1}{4} \, a^{3} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3*x^3,x, algorithm="giac")

[Out]

1/5*b^3*x^5 + 9/14*a*b^2*x^(14/3) + 9/13*a^2*b*x^(13/3) + 1/4*a^3*x^4